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Let AA be set of all real numbers and RR be a relation on AA defined by R={(a,b):1+ab>0}R=\{(a,b): 1+ab>0\}, then relation RR is

Solution

✅ Correct Option: 2

1+a2>01+a^2>0 always, so RR is reflexive, and ab=baab=ba makes it symmetric. For transitivity take a=1a=1, b=0b=0, c=−1c=-1: (1,0)(1,0) and (0,−1)(0,-1) belong to RR, but 1+(1)(−1)=01+(1)(-1)=0 is not positive, so (1,−1)(1,-1) does not.

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